The probability function f associated with a continuous random variable X has the form f(x) = ax^2 + bx (0 ≤ x ≤ 1). If E(X) = 0.6, find the values of a and b
-
(i) Remember that the total area under f equals 1.
So, ∫(x = 0 to 1) (ax^2 + bx) dx = 1
==> a/3 + b/2 = 1
==> 2a + 3b = 6.
---------------
(ii) Since E(X) = ∫(x = 0 to 1) x * (ax^2 + bx) dx, we have
∫(x = 0 to 1) x * (ax^2 + bx) dx = 0.6
==> ∫(x = 0 to 1) (ax^3 + bx^2) dx = 0.6
==> 3a + 4b = 7.2
Now, solve for a and b:
==> a = -2.4 and b = 3.6.
I hope this helps!
So, ∫(x = 0 to 1) (ax^2 + bx) dx = 1
==> a/3 + b/2 = 1
==> 2a + 3b = 6.
---------------
(ii) Since E(X) = ∫(x = 0 to 1) x * (ax^2 + bx) dx, we have
∫(x = 0 to 1) x * (ax^2 + bx) dx = 0.6
==> ∫(x = 0 to 1) (ax^3 + bx^2) dx = 0.6
==> 3a + 4b = 7.2
Now, solve for a and b:
==> a = -2.4 and b = 3.6.
I hope this helps!
-
Since f(x) is a probability density function, then area under f(x) = 1
∫₀¹ (ax² + bx) = 1
(1/3 ax³ + 1/2 bx²) |₀¹ = 1
1/3 a + 1/2 b = 1
2a + 3b = 6
E(X) = 0.6
∫₀¹ x f(x) dx = 0.6
∫₀¹ x (ax² + bx) dx = 0.6
∫₀¹ (ax³ + bx²) dx = 0.6
(1/4 ax⁴ + 1/3 bx³) |₀¹ = 0.6
1/4 a + 1/3 b = 6/10 -------> multiply both sides by 60
15a + 20b = 36
Now solve system of equations:
a = −12/5 = −2.4
b = 18/5 = 3.6
f(x) = −2.4x² + 3.6x (0 ≤ x ≤ 1)
∫₀¹ (ax² + bx) = 1
(1/3 ax³ + 1/2 bx²) |₀¹ = 1
1/3 a + 1/2 b = 1
2a + 3b = 6
E(X) = 0.6
∫₀¹ x f(x) dx = 0.6
∫₀¹ x (ax² + bx) dx = 0.6
∫₀¹ (ax³ + bx²) dx = 0.6
(1/4 ax⁴ + 1/3 bx³) |₀¹ = 0.6
1/4 a + 1/3 b = 6/10 -------> multiply both sides by 60
15a + 20b = 36
Now solve system of equations:
a = −12/5 = −2.4
b = 18/5 = 3.6
f(x) = −2.4x² + 3.6x (0 ≤ x ≤ 1)